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The lattice parameter of an element stabilized in a fcc structure is 4.04 Å. The atomic radius of the element is:
  • a)
    2.86 Å
  • b)
    1.43 Å
  • c)
    4.29 Å
  • d)
    5.72 Å
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The lattice parameter of an element stabilized in a fcc structure is 4...
Å. This means that the distance between adjacent atoms in the crystal lattice is 4.04 Å. The fcc structure has a face-centered cubic arrangement, where each atom is located at the center of a cube and at the center of each face of the cube. The lattice parameter is the length of the edge of the cube. The fcc structure is the most efficient packing arrangement for spheres, and is commonly found in metals such as copper, silver, and gold.
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Community Answer
The lattice parameter of an element stabilized in a fcc structure is 4...
R=a/(2√2)
so 4/(2√2)=1.43
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The lattice parameter of an element stabilized in a fcc structure is 4.04 Å. The atomic radius of the element is:a)2.86 Åb)1.43 Åc)4.29 Åd)5.72 ÅCorrect answer is option 'B'. Can you explain this answer?
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